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Quadratic Equation Problem

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Quadratic Equation Problem

by [email protected] » Tue Jan 01, 2013 11:08 am
If \frac{3}{x+3}+\frac{4}{x+4}=\frac{2}{x+2}+\frac{5}{x+5}, then the value of x is



A. \frac {13}{4}


B. -\frac {5}{2}


C. -\frac {7}{2}


D. \frac {9}{2}


E. \frac {17}{2}

Can someone kindly help me in answering this question. The source of this question is Jumbo Tests.

Thanks in advance.
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by Jim@StratusPrep » Wed Jan 09, 2013 6:50 am
It is a little hard to understand the notations. Can you rewrite or potentially use an image?
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by jitsy » Fri Jul 19, 2013 7:15 am
Could anyone solve this ques. The only thing I can figure out is that in the ques itself, in a way, 3 and 4 have been written as 2+1 and 5-1 which becomes 2+5 (if you know what I mean). Experts?
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by [email protected] » Fri Jul 19, 2013 5:25 pm
Hi everyone,

I think that the question is supposed to be this:

3/(x+3) + 4/(x+4) = 2/(x+2) + 5/(x+5) x = ?

13/4
-5/2
-7/2
9/2
17/2

I've come to a solution by using the following math steps:

First, set a common denominator for the left and a separate common denominator for the right. You'll notice an interesting PATTERN.

3(x+4) + 4(x+3) 2(x+5) + 5(x+2)
------------- = ----------------
(x+3)(x+4) (x+2)(x+5)


7x + 24 7x + 20
------------- = --------------
x^2 + 7x + 12 x^2 + 7x + 10

Notice how the numerators and denominators have several things in common? That means when you cross-multiply, many of the terms will cancel out.

After cross-multiplying, simplifying, and canceling out terms, you'll get....

4x^2 + 14x = 0

2x(2x + 7) = 0

x = 0, -7/2

Final Answer: C

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by pitta.deepak » Tue Jul 30, 2013 1:23 am
Step 1: 3/(x+3) + 4/(x+4) = 2/(x+2) + 5/(x+5)

Step 2: 3/(x+3) - 2/(x+2) = 5/(x+5)- 4/(x+4)

Step 3: x/(x+3)(x+2) = x/(x+5)(x+4)

Step 4: (x+3)(x+2) = (x+5)(x+4)

Step 5: Simplify to get x= -7/2

Step 2 is crucial to solve it faster.
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